Chapter 3

CLB 10703 Sept 2014

CHAPTER 3

Chapter 3A :

1. Introduction to Equilibrium Concept

2. Chemical Equilibrium

3. Equilibrium Constant

4. Heterogeneous & Homogeneous Equilibrium

5. Reaction Quotient

6. Calculating Equilibrium Concentrations

7. Le Chteliers Principle

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Equilibrium???

OPEN SYSTEM

CLOSED SYSTEM

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A state in which there are no observable changes as time goes by as long as the system remains undisturbed.

Chemical equilibrium is achieved when: Occur only in a closed system. The rates of forward & reverse reactions are equal. the concentration of the reactants and the

concentration of the products remain constant.

Physical Equilibrium

a) H2O (l) H2O (g) b) I2 (l) I2 (g)

Equilibrium

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N2O4 (g) 2NO2 (g)

The composition of colorless gas, dinitrogen tetroxide to dark brown gas, nitrogen oxide :

- At the beginning of the rxn, no NO2 reverse rxn does not occur, only production of NO2 will occur.

- After sometime, amount of NO2 , the 2 molecules of NO2 will collide constructively to form N2O4.

So, the forward chemical rxn has an opposing rxn that will with the in product formation.

Chemical Equilibrium

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N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

equilibrium

equilibrium

equilibrium

Chemical Equilibrium

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constant

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The Equilibrium Constant, K indicates a rxn is favored to the products or reactants & can be used to calculate the quantity of the reactants and the products present at equilibrium.

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]b

K = Equilibrium Constant a, b, c, and d = stoichiometric coefficients A & B = reactants C & D = products

Law of Mass Action

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N2O4 (g) 2NO2 (g)

= 4.63 x 10-3 K = [NO2]

2 [N2O4]

K >> 1

K

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Heterogeneous & Homogeneous Equilibrium

Homogeneous equilibrium: applies to reactions in which all reacting species are in the same phase.

eg: solution or gas phase

all in gas phase

Heterogeneous equilibrium: When one or more reactants or products are in the different phases.

N2 (g) + 3H2 (g) 2NH3 (g)

CaCO3 (s) CaO (s) + CO2 (g)

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Homogenous Equilibrium

Kc = [NO2]

2

[N2O4]

N2O4 (g) 2NO2 (g)

Kp = NO2

P2

N2O4 P

In most cases,

Kc Kp

aA (g) + bB (g) cC (g) + dD (g)

Kp = Kc(RT)Dn

n = moles of gaseous products moles of gaseous reactants

= (c + d) (a + b)

Kc = Equilibrium constant in terms of molar conc.

Kp = Equilibrium constant in terms of partial pressure.

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CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O

+ (aq)

Kc = [CH3COO

-][H3O+]

[CH3COOH][H2O] [H2O] = constant

Kc = [CH3COO

-][H3O+]

[CH3COOH] = Kc [H2O]

General practice not to include UNIT for the equilibrium constant.

Homogenous Equilibrium

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The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 74C are [CO] = 0.012 M, [Cl2] = 0.054 M & [COCl2] = 0.14 M. Calculate the equilibrium constants, Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)

Kc = [COCl2]

[CO][Cl2] =

0.14 0.012 x 0.054

= 220

Kp = Kc(RT)n

n = 1 2 = -1 R = 0.0821 T = 273 + 74 = 347

Kp = 220 x (0.0821 x 347)-1 = 7.7

Example :

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The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO2 = 0.400 atm and PNO = 0.270 atm?

2NO2 (g) 2NO (g) + O2 (g)

Kp = 2

P NO PO 2

P NO 2

2

PO 2 = Kp PNO

2 2

P NO 2

PO 2 = 158 x (0.400)2/(0.270)2 = 347 atm

Example :

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Equilibrium constant, Kc for reaction below at 2300C is 2.50 x 1012.

2O3 (g) 3O2 (g)

What is the Kp value at the same temperature?

Kp = Kc(RT)n

Kp = 2.50 x 1012 (0.0821 x 2573) Dn

Dn = 3 2 = 1

Kp = 5.28 x 1014

Example :

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Heterogenous equilibrium applied to reactions in which reactants & products are in different phases.


Kc = [CaO][CO2]

[CaCO3]

[CaCO3] = constant [CaO] = constant

Kc = [CO2] = Kc x [CaCO3]

[CaO] Kp = PCO 2

The concentrations of solids and pure liquids are not included in the expression for the equilibrium constant.

Heterogenous Equilibrium

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PCO 2 = Kp


PCO 2 does not depend on the amount of CaCO3 or CaO

(a) (b)

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Consider the following equilibrium at 295 K:

The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction.

NH4HS (s) NH3 (g) + H2S (g)

Kp = P NH3 H2S P = 0.265 x 0.265 = 0.0702

Kp = Kc(RT)Dn

Kc = Kp(RT)-Dn

Dn = 2 0 = 2 T = 295 K

Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

re-arrange

Example :

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N2O4 (g) 2NO2 (g)

= 4.63 x 10-3 K = [NO2]

2

[N2O4]

2NO2 (g) N2O4 (g)

K = [N2O4]

[NO2]2

= 1 K

= 216

When the equation for a reversible reaction is written in the opposite

direction, the equilibrium constant will become the reciprocal of the original

equilibrium constant.

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A + B C + D

C + D E + F

A + B E + F

Kc = [C][D]

[A][B]

Kc = [E][F]

[C][D]

[E][F]

[A][B] Kc =

Kc

Kc

Kc

Kc = Kc Kc x

If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the

individual reactions.

K for Multiple Reactions

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The conc. of reacting species in condensed phase : use M.

For gaseous phase : use M or atm.

The conc. of pure solids, pure liquids & solvents do not

appear in the equilibrium constant expressions.

The equilibrium constant is a dimensionless quantity.

In quoting a value for the equilibrium constant, must specify

the balanced equation & the temperature.

If a reaction can be expressed as a sum of two or more

reactions, the equilibrium constant for the overall reaction is

given by the product of the equilibrium constants of the

individual reactions.

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A + 2B AB2 Kf

Kr

ratef = kf [A][B]2

rater = kr [AB2]

At Equilibrium: rate forward = rate reverse

kf [A][B]2 = kr [AB2]

kf

kr

[AB2]

[A][B]2 = Kc =

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A number equal to the ratio of product concentration to

reactants concentrations, each raised to the power of its

stoichiometric coefficient at some point other than

equilibrium.

Reaction Quotient (Qc)

The reaction quotient (Qc) is

calculated by substituting the

initial concentrations of the

reactants and products into

the equilibrium constant (Kc)

expression.

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Qc > Kc system proceeds from right to left to reach equilibrium Qc = Kc the system is at equilibrium Qc < Kc system proceeds from left to right to reach equilibrium

Reaction Quotient (Qc)

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1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.

2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, K then solve for x.

3. Having solved for x, calculate the equilibrium concentrations of all species.

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At 1280C, the reaction below has a equilibrium constant, Kc = 1.1 x 10

-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Let x be the change in concentration of Br2

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

Calculating Equilibrium Concentrations

Br2 (g) 2Br (g)

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[Br]2

[Br2] Kc = =

(0.012 + 2x)2

0.063 - x = 1.1 x 10-3

Solve for x :

4x2 + 0.048x + 0.000144 = 0.0000693 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

ax2 + bx + c =0

-b b2 4ac 2a

x =

x2 = - 0.00178 x1 = - 0.0105

Use :

Br2 (g) 2Br (g)

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Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M

At equilibrium, [Br2] = 0.062 x = 0.0648 M

x2 = -0.00178 x1 = -0.0105

Br2 (g) 2Br (g)

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At a given temperature, 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium, 0.20 mole of NH3 was present. Calculate Kc for the reaction.

N2 (g) + 3H2 (g) 2NH3 (g)

Kc = [NH3]

2

[N2] [H2]3

1. State the chemical equation & balance it

2. Write the equilibrium constant.

Example :

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Initial (M)

Change (M)

0.8 0.9 0

- 0.1 - 0.3 + 0.2

0.7 0.6 0.2 Equilibrium (M)

N2 (g) + 3H2 (g) 2NH3 (g)

Kc = = 0.265 [NH3]

2

[N2][H2]3

3. Construct an ICE table..

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Calculate Kc for reaction below if at equilibrium [H2] = 0.2M, [I2] = 0.02 M and [HI] = 1.6 M at 445C.

H2 (g) + I2 (g) 2HI (g)

Kc = = [HI]2

[H2] [I2]

= 64.0 (1.6)2

(0.2) (0.2)

If in the other trial, 4.0 M of HI was added into the above reaction. Calculate the [H2 ],

[I2] & [HI] after reached the equilibrium.

Example :

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2HI (g) H2 (g) + I2 (g)

Initial (M)

Change (M)

Equilibrium (M)

4 0 0

-2x x x

4 2x x x

= Kc 1

Kc

[H2] [I2] Kc'= =

[HI]2 =

1

64

x2

(4 2x)2

= 1

64

x2

(4 2x)2

[H2] = [I2] = 0.4 M [HI] = 3.2 M

x = 0.4

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If an external stress is applied to a system at equilibrium,

the system adjusts in such a way that the

stress is partially offset as the system

reaches a new equilibrium position.

Le Chteliers Principle

Changes in Concentration

N2 (g) + 3H2 (g) 2NH3 (g)

Add NH3

Equilibrium shifts backward to offset

the stress

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Changes in Concentration

Change Shifts the

Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

aA + bB cC + dD


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Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas

Decrease volume Increase volume Side with most moles of gas

Side with fewest moles of gas


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Changes in Temperature

Change Exothermic Rxn

temperature K decreases

temperature K increases

Endothermic Rxn

K increases

K decreases

Adding a Catalyst

1. does not change K 2. does not shift the position of an equilibrium system 3. The system will reach equilibrium sooner


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uncatalyzed

Catalyst lowers Ea for both forward & reverse reactions.

Adding a Catalyst

catalyzed

A catalyst speeds up both the forward and reverse reactions by exactly the same amount.!

Catalyst does not change the K or shift the equilibrium!!

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CHAPTER 3

Le Chteliers Principle - summary

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Chapter 3

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